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# HW 15 - johnson(mjj622 HW 15 Coker(56625 This print-out...

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johnson (mjj622) – HW 15 – Coker – (56625) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This assignment covers the first chapter on wave motion, Ch. 16, which emphasizes trans- verse waves, the wave equation, wave power, reflection, superposition, standing waves and group speed. 001 10.0 points A transverse wave on a very long string is described by y ( x, t ) = (2 cm) sin bracketleftbig( 1 . 6 m - 1 ) x - ( 50 s - 1 ) t bracketrightbig What is the acceleration of the point at x = 0 at t = 3 π 4 seconds? 1. 50 m/s 2 2. - 10 cm/s 2 3. - 200 cm/s 2 4. - 50 m/s 2 correct 5. 2 cm/s 2 6. Zero; this is a standing wave. 7. 200 cm/s 2 8. - 62 . 5 cm/s 2 9. 10 cm/s 2 10. 1.6 m/s 2 Explanation: For y ( x, t ) = A sin( k x - ω t ), the accelera- tion of any point on the string is a y ( x, t ) = 2 y ∂t 2 = - A ω 2 sin ( k x - ω t ) . sin( - θ ) = - sin θ , so at x = 0 a y (0 , t ) = - A ω 2 sin ( - ω t ) = A ω 2 sin ( ω t ) , and for t = 3 π 4 , a y = (2 cm) ( 50 s - 1 ) 2 sin bracketleftbigg ( 50 s - 1 ) parenleftbigg 3 π 4 s parenrightbiggbracketrightbigg = ( 5000 cm / s 2 ) sin parenleftbigg 75 π 2 parenrightbigg = ( 5000 cm / s 2 ) sin parenleftbigg 36 π + 3 π 2 parenrightbigg = ( 5000 cm / s 2 ) sin parenleftbigg 3 π 2 parenrightbigg = ( 5000 cm / s 2 ) ( - 1) × 1 m 100 cm = - 50 m / s 2 . 002 10.0 points A harmonic wave y = A sin[ k x - ω t - φ ] , where A = 1 m, k has units of m - 1 , ω has units of s - 1 , and φ has units of radians, is plotted in the diagram below. +1 - 1 A (meters) x (meters) 2 4 6 At the time t = 0 Which wave function corresponds best to the diagram? 1. y = A sin bracketleftbigg parenleftbigg 2 π 15 m parenrightbigg x - ω t - parenleftbigg 1 π 3 parenrightbiggbracketrightbigg 2. y = A sin bracketleftbigg parenleftbigg 2 π 15 m parenrightbigg x - ω t - parenleftbigg 5 π 3 parenrightbiggbracketrightbigg 3. y = A sin bracketleftbigg parenleftbigg 2 π 9 m parenrightbigg x - ω t - parenleftbigg 5 π 3 parenrightbiggbracketrightbigg 4. y = A sin bracketleftbigg parenleftbigg 2 π 9 m parenrightbigg x - ω t - parenleftbigg 4 π 3 parenrightbiggbracketrightbigg

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johnson (mjj622) – HW 15 – Coker – (56625) 2 5. y = A sin bracketleftbigg parenleftbigg 2 π 3 m parenrightbigg x - ω t - parenleftbigg 2 π 3 parenrightbiggbracketrightbigg 6. y = A sin bracketleftbigg parenleftbigg 2 π 15 m parenrightbigg x - ω t - parenleftbigg 2 π 3 parenrightbiggbracketrightbigg 7. y = A sin bracketleftbigg parenleftbigg 2 π 3 m parenrightbigg x - ω t - parenleftbigg 1 π 3 parenrightbiggbracketrightbigg 8. y = A sin bracketleftbigg parenleftbigg 2 π 9 m parenrightbigg x - ω t - parenleftbigg 1 π 3 parenrightbiggbracketrightbigg 9. y = A sin bracketleftbigg parenleftbigg 2 π 3 m parenrightbigg x - ω t - parenleftbigg 5 π 3 parenrightbiggbracketrightbigg 10. y = A sin bracketleftbigg parenleftbigg 2 π 3 m parenrightbigg x - ω t - parenleftbigg 4 π 3 parenrightbiggbracketrightbigg cor- rect Explanation: From the diagram of the wave function the wave-length λ = 3 m (6 horizontal scale divi- sions of 0 . 5 m each, see diagram below).
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HW 15 - johnson(mjj622 HW 15 Coker(56625 This print-out...

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